Countable bounded multisets admit injective sum encodings

Let $B=\max \{2,M+1\}$ and set $\varphi (x)=B^{\nu (x)}$. If $M=0$, the only allowed multiset is empty, so its sum encoding is injective. If $M\ge 1$ and two allowed multisets $S,T$ have equal sums, for each $j\in {\mathbb{N}}_0$ let $d_j$ be the multiplicity in $S$ minus the multiplicity in $T$ of the element with $\nu$-index $j$, and set $d_j=0$ when no element has that index. The sequence $(d_j)$ has finite support and $\mathrm{\mid }{d}_j\mathrm{\mid }\le M$. If some $d_j\mathrm{\ne }0$, let $r$ be its largest such index. Equality of the sums gives $\mathrm{\mid }{d}_r\mathrm{\mid }{B}^r=\mathrm{\mid }{\sum }_{j<r}{d}_j{B}^j\mathrm{\mid }\le M(B^r-1)\mathrm{/}(B-1)\le B^r-1$, contradicting $\mathrm{\mid }{d}_r\mathrm{\mid }{B}^r\ge B^r$. Hence all multiplicities agree and $S=T$.

Suppose $(1+\epsilon )\varphi (c)+{\sum }_{x\in S}\varphi (x)=(1+\epsilon )\varphi (c^{\mathrm{\prime }})+{\sum }_{x\in T}\varphi (x)$. Rearranging gives $A+\epsilon D=0$, where $A=\varphi (c)-\varphi (c^{\mathrm{\prime }})+{\sum }_{x\in S}\varphi (x)-{\sum }_{x\in T}\varphi (x)$ and $D=\varphi (c)-\varphi (c^{\mathrm{\prime }})$ are integers. If $D\mathrm{\ne }0$, then $\epsilon =-A\mathrm{/}D$ is rational, contrary to the hypothesis. Thus $D=0$; since $B\ge 2$ has distinct powers and $\nu$ is injective, $\varphi$ is injective and $c=c^{\mathrm{\prime }}$. The equality then reduces to equality of the neighbor sums; step 1.1 gives $S=T$.